1281. Can Make Palindrome From Substring¶

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1281. Can Make Palindrome from Substring

Medium

You are given a string s and array queries where queries[i] = [lefti, righti, ki]. We may rearrange the substring s[lefti...righti] for each query and then choose up to ki of them to replace with any lowercase English letter.

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return a boolean array answer where answer[i] is the result of the ith query queries[i].

Note that each letter is counted individually for replacement, so if, for example s[lefti...righti] = "aaa", and ki = 2, we can only replace two of the letters. Also, note that no query modifies the initial string s.

Example :

Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
Explanation:
queries[0]: substring = "d", is palidrome.
queries[1]: substring = "bc", is not palidrome.
queries[2]: substring = "abcd", is not palidrome after replacing only 1 character.
queries[3]: substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries[4]: substring = "abcda", could be changed to "abcba" which is palidrome.

Example 2:

Input: s = "lyb", queries = [[0,1,0],[2,2,1]]
Output: [false,true]

Constraints:

• 1 <= s.length, queries.length <= 105
• 0 <= lefti <= righti < s.length
• 0 <= ki <= s.length
• s consists of lowercase English letters.

Solution¶

class Solution {
    private int pref[][];
    public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
        int n = s.length();
        pref = new int[n][26];

        for (int i = 0; i < n; i++) {
            char current = s.charAt(i);
            for (int j = 0; j < 26; j++) 
                if (i - 1 >= 0) 
                    pref[i][j] = pref[i - 1][j];
            pref[i][current - 'a']++;
        }

        List<Boolean> res = new ArrayList<>();
        for (int query[] : queries) {
            int l = query[0], r = query[1], k = query[2];
            int countOdd = 0;
            for (int i = 0; i < 26; i++) {
                int total = pref[r][i];
                if (l - 1 >= 0) 
                    total -= pref[l - 1][i];
                if (total % 2 == 1) 
                    countOdd++;
            }
            if (2 * k >= countOdd - 1) 
                res.add(true);
            else 
                res.add(false);
        }
        return res;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here